Problem Solving and Computing (CSC-103 98S)

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# Resolutions for Problem Set 4

Disclaimer. My resolutions for this set are even "live"r than most. I wrote down my own processes and thoughts as I went, rather than predicting all of the alternatives you might try. Clearly, your resolutions will differ from some of these.

0. Gender-based Group Learning

There are many possible reasons for this. A simple one is that "women are smarter and therefore raise the curve in any group" (yes, there are ways to argue the reverse, but it's more difficulty). Another is that the different groups have significantly different characters and the way men and women behave in different types of groups affects their learning. (You can fill in their behaviors.)

How would we tell? If we make a claim about behavior, we can observe the groups and see if our observations match the claim. Clearly, we would want to observe a wide variety of groups in a wide variety of situations.

I started my consideration of this problem by drawing a timeline.

```| childhood | youth | unmarried | childless | parent |
+----------------------------------------------------+
|son dies
```

I then reread the problem and realized that some of the numbers didn't make sense, particularly as I listened to my quite able students discussing the problem. I checked my typing and realized that I had mistyped the amount of time in childhood. Sorry ...

Okay, on to defining variables.

• D will be the age at which the woman died.
• C will be the amount of time spent in childhood.
• Y will be the amount of time spent in youth.
• U will be the amount of time spent unmarried.
• M will be the amount of time spent married by childless.
• P will be the amount of time spent as a parent.
• S will be the age of her son when she dies.

Yuck, that's a lot of variables. On to the fun equations.

• C = 1/6 * D (one sixth of her life in childhood)
• Y = 1/12 * D (one twelfth of her life in youth)
• U = 1/7 * D (one seventh single) Aha! I have an alternative idea I'd like to follow up on. I'll come back to the equations later.

Okay, I'll assume that the person who wrote this problem didn't want to work with fractional ages. Hence, her age has to be a multiple of both seven and twelve. If I had to guess, I'd guess 84, which is 7 times twelve. How can I check this? It might help to have the equations, so I'll continue with those equations.

• S = P - 4 (her son died four years before she did, so his age is four years less than she was a "parent")
• S = D / 2 (her son was half her age when he died)
• M = 5 (she was married for five years before she had a child)
• C + Y + U + M + P = D (the sum of the parts of her life is equal to the span of her life)

First, I'll resolve the two equations using S so that I can eliminate S. Now, we know that D/2 = P-4 or that P = D/2+4 (which is a good form, since everything else is in terms of D).

Next, I'll solve that big sum for D (which shouldn't be so bad, since the other equations are in terms of D).

• (1/6)*D + (1/12)*D + (1/7)*D + 5 + (1/2)*D + 4 = D

Isn't that ugly? Well, I guess I'll find a common denominator. Hey! It's 84.

• (14/84)*D + (7/84)*D + (12/84)*D + 5 + (42/84)*D + 4 = D

Okay, now I can simplify a little.

• [(14+7+12+42)/84]*D + 9 = D
• (75/84)*D + 9 = D
• 9 = D - (75/84)*D
• 9 = (84/84)*D - (75/84)*D
• 9 = 9/84*D
• D = 84

Not much of a surprise, but a useful answer.

Check.

• 14 years in childhood.
• 7 years in youth. I'm not sure that I'd call the ages between 14 and 21 youth, but okay.
• 12 years single. Hmmm ... she was 33 before she married.
• 5 years childless. Hmmm ... she was 38 when she had children.
• She died 46 years later (at 82), so her son was 42 when he died.

Reflect. That was a lot of work for a not very intersting (and potentially obvious) answer. As in some previous word problems, I found it more interesting to reflect on the "hidden" values in the problem and solve with them, rather than to work out the formulae. However, I did not a problem in the problem when I tried to work out the formulae.

2. Three More Generations

Oh boy, another one of these aging problems. Well, let's see.

• M is my mother's age.
• I is my age.
• D is my daughter's age.

And formulae

• My mother is twice as old as I am.
• M = 2*I
• The ratio between my mother's age and my age is half the ration between my age and my daughter's age.
• M/I = 1/2 * I/D
• My mother is eight times as old as my daughter.
• M = 8*D

Let's see, we'll express I in terms of M in the first equation and D in terms of M in the last equation and then plug into the middle equation.

• I = M/2
• D = M/8
• M/(M/2) = 1/2 * (M/2)/(M/8)

Simplifying

• 2 = 1/2 * (1/2)/(1/8)
• 2 = 1/2 * 8/2
• 2 = 2

Hmmm ... Not a very useful result. Stuck!

Let's think about those initial statements again. Aha! Wait a minute, if my mother is twice as old as I am and eight times as old as my daughter, then I must be four times as old as my daughter. That means that I already knew that the ratio of her age to my age was half the ration of my age to my daughter's age! I only have two useful equations and three unknowns, so the problem isn't solvable for particular values.

In fact, we can choose a large number of ages that meet these criteria.

• We could use M=40, I=20, D=5
• We could use M=48, I=24, D=6
• We could use a lot of similar things, especially if we allow fractions.

Check. Yes, all of those examples meet the criteria.

Reflect. Although this was similar to a number of "how old are we" problems, the author of this problem (SamR) either didn't try to ensure that the equations were independent or purposefully wrote the question so that they didn't give sufficient information (okay, I wrote them, I know that it's the latter). Did I learn something from this: yes, sometimes the same technique does not work the same way for all similar problems. I also learned that the phrasing of problems is important.

3. Selling Candy

Oh boy, more numeric problems. 9,000 have nuts and/or chocolate. We'll divide the world once more.

• C: chocolate only
• N: nuts only
• Y: nuts and chocolate (for yech or yum, depending on your perspective)

On to equations:

• C+Y = 6500 (6500 have chocolate)
• N+Y = 7200 (7200 have nuts)
• C+N+Y = 9000 (1000 of 10000 have neither)

Plug in 7200 for C+Y in the third equation and we get

• C + 7200 = 9000
• C = 1800

Plug in 1800 for C in the first equation and we get

• 1800+Y = 6500
• Y = 4700

Plug in 4000 for Y in the second equation and we get

• N+4700 = 7200
• N = 2500

Check.

• C+Y = 1800 + 4700 = 6500
• N+Y = 2500 + 4700 = 7200
• C+N+Y = 1800 + 2500 + 4700 = 9000

Reflect This was very similar to the sandwiches problem from problem set 1.

4. The Fake Coin

We did this as a group for nine coins.

5. Campus Bikes

In order to understand the long term effect, we should choose a number of years that will give us even numbers. Since we have seen issues of ten years and fifteen years, we might as well look at thirty years. Does it matter how many bikes we buy? Apparently not, since we'll just multiply by the number of bikes. Does it matter how much money we start out with? If we start out with too little, it's a problem. However, the endowment is big enough that we don't need to worry about it. If we have some extra money in both cases, that's okay, too. We'll start with \$10,000, since that's obviously enough to purchase a few generations of bikes.

Case one: no Zoomcoat, no interest, thirty years. In year 0, we buy one Zoomster, leaving us with \$8,900. In year 10, we sell that Zoomster for \$100, giving us \$9,000. We then buy another bike, leaving us with \$7,900. In year 20, we sell that Zoomster for \$100, giving us \$8,000. We then buy another one for \$1,100, leaving us with \$6,900. In year 30, we sell that for \$100, leaving us with \$7,000.

Case two: Zoomcoat, no interest, thirty years. In year 0, we buy one Zoomster and Zoomcoat, leaving us with \$8,600. In year fifteen, we buy another Zoomster and Zoomcoat, leaving us with \$7,200. This looks like a win.

Observe. I note that \$3,000 would have been enough to start with. Will that make analysis of the problem easier? Perhaps, but not enough so to warrant many changes.

Reflect. We have an answer in a no-interest world, but it seems as if we might make more money by investing that \$300, especially since some of it will be invested for a long time. How much do I make? In ten years, I make principal*(1.05)^10 or approximately principal*1.63. In fifteen years, I make principal*(1.05)^15 or approximately principal*2.08.

Case three: no Zoomcoat, interest, thirty years. In year 0, we buy one Zoomster, leaving us with \$8,900. In year 10, we have collected approximately \$5,607 in compound interest, giving us \$14,507. We also sell the Zoomster for \$100, giving us \$14,607. We buy another Zoomster for \$1,100, leaving us with \$13,507. In year 20, we have collected approximately \$8,509 in compound interest, giving us \$22,016. We sell the Zoomster for \$100, giving us \$22,116. We then buy another for \$1,110, leaving us with \$21,006. Finally, in year 30, we have collected approximately \$13,233 in interest, giving us \$34,239. We sell the remaining Zoomster for \$100, and end up with \$34,339. Not a bad chunk of change.

Case four: Zoomcoat, interest, thirty years. In year 0, we buy one Zoomster with Zoomcoat for \$1,400, leaving us with \$8,600. In year 15, we have collected approximately \$9,288 in compound interest, giving us \$17,888. We then buy another Zoomster with Zoomcoat, leaving us with \$16,488. In year 30, we have collected approximately \$17,807 in compound interest, giving us \$34,295.

Check. Um, I did lots of rounding in this problem, and it's likely that the bank would, too. I'm not sure whether or not my answer is correct. If pressed, I'd choose to do without Zoomcoat, but wouldn't argue strenuously in either direction. I might also want to try the same problem with different starting amounts. It seems with a higher interest rate I might do a little bit better.

Reflect. Not a very satisfactory solution, although I found the problem interesting as it was a little bit more "real world" than some of the other problems. I did notice that even at 5%, you can make a lot of money on your money after thirty years.

Aha!. There does seem to be another way to think about the problem that helps us ignore some of the problematic issues. If I don't buy Zoomcoat then the amount extra I have to invest depends on which year it is. That is, rather than considering amount remaining in my account, I'll consider the extra amount I have for each strategy. I might be able to use this to compute something useful.

• In year zero: We have \$300 extra to invest if we don't buy Zoomcoat.
• In year ten: We have \$700 extra to invest if we do buy Zoomcoat (we haven't yet bought a second bike; we would have spent \$2100 on the first two bikes).
• In year fifteen: We have \$700 extra to invest if we don't buy Zoomcoat (we've spent \$2100 for the first uncoated bikes and \$2800 for the coated bikes).
• In year twenty. We have \$300 extra to invest if we do buy Zoomcoat (we've spent \$3100 for the first three uncoated bikes and \$2800 for the coated bikes).

Now, how does this extra cash affect our income?

In the "no Zoomcoat" case.

• \$300 extra for 30 years (years zero through thirty)
• \$300*(1.05)^30
• \$700 extra for 15 years (years fifteen through thirty)
• \$700*(1.05)^15
• \$100 refund at the end
• \$100

In the "Zoomcoat" case.

• \$700 extra for 20 years (years ten through thirty)
• \$700*(1.05)^20
• \$300 extra for 10 years (years twenty through thirty)
• \$300*(1.05)^10

Does this help? At least I can come up with numbers that better describe the differential.

• No Zoomcoat
• \$300*(1.05)^30 + \$700*(1.05)^15 + \$100
• \$1,296 + \$1,455 + \$100
• \$2,851.
• Zoomcoat
• \$700*(1.05)^20 + \$300*(1.05)^10
• \$1,857 + \$488
• \$2,435

It does seem that skipping the Zoomcoat is better.

Check. If my reasoning is sound, the differences between the numbers in both cases should be about the same. \$2,851 - \$2,435 = \$416. \$34,339 - \$34,295 = \$44. Um. That's not good.

Stuck!.

6. Differential Taxes

This seems like a good problem in which to write formulae. Let I be my untaxed (gross) income. My real (net) income will be I * (1-(I/5000)/100). or (-1/500,000)*I^2 + I. That's a quadratic equation describing a parabola that "points downwards". I know that parabolas are symmetrical, so the high point on this parabola will be midway between the two zeroes of the function.

In A*I^2 + B*I + C form,

• A = -1/500,000
• B = 1
• C = 0

I could then find the zeroes of the function by using the legendary formula (b*b +- 4ac)/2a. However, since this is such a simple function, I can also write it as I((-1/500,000)I + 1). This is 0 when I is 0 and when I is 500,000. Hence, I take home the most money when my gross income is midway between those two values. I should strive to earn \$250,000.

Check. I'll try a few examples to see if my answer seems reasonable. If I make \$250,000, I take home 50% of \$250,000, or \$125,000. If I make \$200,000, I take home 60% or \$120,000. If I make \$300,000, I take home 40% or \$120,000.

Reflect. What does this do to Bill Gates? More realistically, it suggests that this taxation system caps people's real income, at a number that will feel high to many Americans and low to a few Americans. It might be fun to look up incomes in the census and see what it really does (and whether it would work).

7. The Bricklayers

• A - the number of bricks per hour that the first bricklayer can lay.
• B - the number of bricks per hour that the second bricklayer can lay.
• W - number of bricks in the wall.

The first bricklayer takes 9 hours to build the wall.

• W = 9*A

The second bricklayer takes 10 hours to build the wall.

• W = 10*B

The two bricklayers can build the wall in 5 hours, but they two bricklayers lay 10 fewer bricks per hour.

• W = 5(A+B-10)

Three equations, three unknowns. Looks good. Let's rephrase the first two equations so that we have A in terms of W and B in terms of W.

• A = W/9 (the first bricklayer lays 1/9 of the wall in one hour)
• B = W/10 (the second bricklayer lays 1/10 of the wall in one hour)

We can now work with the third.

• W = 5(W/9 + W/10 - 10)
• W = 5*(10W/90 + 9W/90 - 10)
• W = 5*(19W/90 - 10)
• W = 85W/90 - 50
• 5/90W = -50
• W = -900

Stuck! That's a pretty meaningless result. How can a wall have negative bricks?

Check. Okay, it appears that I can't do math. 5*19 = 95, rather than 85. Let's try again.

• W = 95W/90 - 50
• 5W/90 = 50
• W = 900
• A = 100 bricks per hour
• B = 90 bricks per hour

Check.

• In nine hours, the first bricklayer lays 900 bricks.
• In ten hours, the second bricklayer lays 900 bricks.
• Each hour, they lay 100+90-10 = 180 bricks. In five hours, they can lay 900 bricks.

Reflect. I'm clearly too confident in my "simple math" skills and pick answers without reflecting sufficiently much. This is yet another "determine variables and formulae and solve" problem. It was a little bit more interesting in that the variables I picked had to do with rates, but that didn't seem to be a big difference.

8. The City Gates

This is one of those "what's really being asked" questions. Clearly, the person who wrote it was trying to rephrase the "St. Ives" question. In one sense, this version doesn't give us enough data. How many were at the city gates? Well, we only know about the people who were described in the question.

This seems to be another one in which it helps to have variables.

• M - number of mates
• S - number of sacks
• C - number of cats
• K - number of kits

The total number of "things" at the city gates is 1 + 1 + M + S + C + K. The first one is "me". The next one is "someone".

• M = 7
• S = 7 * M = 7 * 7 = 49
• C = 7 * S = 7 * 49 = 343
• K = 7 * C = 7 * 343 = 2401

That totals 2802.

Check. Should I be counting sacks? It does say "kits, cats, sacks, and mates. Is there any potential tricks in the problem. Well, it is possible that the mates shared the sacks (it doesn't say "each mate had seven different sacks"), so it could be possible that there are only seven sacks. For example, if I heard "I just met Sam and Michelle. Each of them has a son. How many son's did they have?" It's possible that they have two. It's also possible that they have one. Clearly, this problem is phrased somewhat confusingly.

Reflect. I've always found "St. Ives" a little bit silly. This one has less of a trick, but could also have hidden tricks. "It depends" would be a reasonable answer. "Who cares" would also be a reasonable answer.

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